Problem: Solve for $x$ and $y$ using substitution. ${5x-4y = -10}$ ${x = -y+7}$
Since $x$ has already been solved for, substitute $-y+7$ for $x$ in the first equation. ${5}{(-y+7)}{- 4y = -10}$ Simplify and solve for $y$ $-5y+35 - 4y = -10$ $-9y+35 = -10$ $-9y+35{-35} = -10{-35}$ $-9y = -45$ $\dfrac{-9y}{{-9}} = \dfrac{-45}{{-9}}$ ${y = 5}$ Now that you know ${y = 5}$ , plug it back into $\thinspace {x = -y+7}\thinspace$ to find $x$ ${x = -}{(5)}{ + 7}$ $x = -5 + 7$ ${x = 2}$ You can also plug ${y = 5}$ into $\thinspace {5x-4y = -10}\thinspace$ and get the same answer for $x$ : ${5x - 4}{(5)}{= -10}$ ${x = 2}$